# Isomorphism between Topological Groups.

Let $$G$$ is an abelian profinite group and $$G=\varprojlim G_i$$ (all $$G_i$$ are finite).

Then why $$\hom(\varprojlim G_i,\mathbb{Q}/\mathbb{Z}) = \varinjlim\hom(G_i,\mathbb{Q}/\mathbb{Z})$$ as topological groups.

• That's not true even as groups, you probably confused $\varinjlim$ and $\varprojlim$ – Max Jul 21 at 16:37
• I may be wrong. Can you check profinite groups by Luis Ribes(Lemma 2.9.3)? – math Jul 21 at 19:31
• check the answer(math.stackexchange.com/a/127291/326275). – math Jul 21 at 19:58
• I assume that by $\operatorname{hom}$, you mean the group of continuous homomorphism, isn't it ? Then look for the compact subgroups of $\mathbb{Q/Z}$. Or directly, show that a continuous homomorphism $G\to\mathbb{Q/Z}$ has an open kernel. – Roland Jul 21 at 20:00
• yes, a group of continuous homomorphism. – math Jul 21 at 20:07